Calculus: Early Transcendentals 8th Edition

by
Stewart, James

Answer

$$\int\cos^{-1} xdx=x\cos^{-1}x-\sqrt{1-x^2}+C$$

Work Step by Step

$$A=\int\cos^{-1} xdx$$
When there is only one element like in this case, we would choose $u=\cos^{-1}x$ and $dv=dx$
For $u=\cos^{-1}x$, we have $du=-\frac{1}{\sqrt{1-x^2}}dx$
For $dv=dx$, $v=x$
Apply Integration by Parts to A, we have $$A=uv−\int vdu$$ $$A=x\cos^{-1}x-\int \frac{-x}{\sqrt{1-x^2}}dx$$ $$$A=x\cos^{-1}x+\int \frac{x}{\sqrt{1-x^2}}dx$$
We now apply the Substitution Rule.
Take $z=1-x^2$, then we have $dz=(1-x^2)'dx=-2xdx$. Therefore, $xdx=-\frac{1}{2}dz$.
Also, $\sqrt{1-x^2}=\sqrt z=z^{1/2}$ $$A=x\cos^{-1}x+\int-\frac{1}{2}\frac{1}{z^{1/2}}dz$$ $$A=x\cos^{-1}x-\frac{1}{2}\int z^{-1/2}dz$$ $$A=x\cos^{-1}x-\frac{1}{2}\frac{z^{1/2}}{\frac{1}{2}}+C$$ $$A=x\cos^{-1}x-\sqrt z+C$$ $$A=x\cos^{-1}x-\sqrt{1-x^2}+C$$